Is ${401133}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {401133}= &&{4}\cdot100000+ \\&&{0}\cdot10000+ \\&&{1}\cdot1000+ \\&&{1}\cdot100+ \\&&{3}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {401133}= &&{4}(99999+1)+ \\&&{0}(9999+1)+ \\&&{1}(999+1)+ \\&&{1}(99+1)+ \\&&{3}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {401133}= &&\gray{4\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {4}+{0}+{1}+{1}+{3}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${401133}$ is divisible by $3$ if ${ 4}+{0}+{1}+{1}+{3}+{3}$ is divisible by $3$ Add the digits of ${401133}$ $ {4}+{0}+{1}+{1}+{3}+{3} = {12} $ If ${12}$ is divisible by $3$ , then ${401133}$ must also be divisible by $3$ ${12}$ is divisible by $3$, therefore ${401133}$ must also be divisible by $3$.